QuestionDetermine the number of terms $n$ in the geometr 3) $4+12+36+108 \ldots, S_{n}=39364$

Studdy Solution

STEP 1

1. The sequence is a geometric series.
2. The first term $a$ of the series is 4.
3. The common ratio $r$ can be determined from the series.
4. The sum of the first $n$ terms $S_n$ is given as 39364.

STEP 2

1. Determine the common ratio $r$ .
2. Use the formula for the sum of a geometric series to find $n$ .

STEP 3

Determine the common ratio $r$ by dividing the second term by the first term:
$r = \frac{12}{4} = 3$

STEP 4

Use the formula for the sum of the first $n$ terms of a geometric series:
$S_n = a \frac{r^n - 1}{r - 1}$
Given $S_n = 39364$ , $a = 4$ , and $r = 3$ , substitute these values into the formula:
$39364 = 4 \frac{3^n - 1}{3 - 1}$
Simplify the equation:
$39364 = 2 (3^n - 1)$
Divide both sides by 2:
$19682 = 3^n - 1$
Add 1 to both sides:
$19683 = 3^n$

STEP 5

Solve for $n$ by recognizing that $19683$ is a power of 3. Calculate:
$3^9 = 19683$
Thus, $n = 9$ .
The number of terms $n$ is:
$\boxed{9}$

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